What is the derivative of (x-1/(2xsqrt(x))) ?

The derivative of (x-1/(2xsqrt(x))) is (-x+3)/(4x^{5/2)}

What is the first derivative of (x-1/(2xsqrt(x))) ?

The first derivative of (x-1/(2xsqrt(x))) is (-x+3)/(4x^{5/2)}

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\frac{d}{d x} (\frac{x - 1}{2 x x})

\frac{- x + 3}{4 x ^{\frac{5}{2}}}

Solution steps

\frac{d}{d x} (\frac{x - 1}{2 x x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d x} (\frac{x - 1}{x x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( x - 1 ) x x - \frac{d}{d x} ( x x ) ( x - 1 )}{( x x ) ^{2}}

\frac{d}{d x} (x - 1) = 1

\frac{d}{d x} (x x) = \frac{3 x}{2}

= \frac{1}{2} \cdot \frac{1 \cdot x x - \frac{3 x}{2} ( x - 1 )}{( x x ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{1 \cdot x x - \frac{3 x}{2} ( x - 1 )}{( x x ) ^{2}} : \frac{- x + 3}{4 x ^{\frac{5}{2}}}

= \frac{- x + 3}{4 x ^{\frac{5}{2}}}

y^{^{''}} + 16 y = sec (4 t)

f (x) = 10 x

\frac{d y}{d x} = (e^{x + y})

\frac{d y}{d x} - \frac{y}{x} = x sin (x)

x \to 1 + lim (ln (x^{3} - 1))

What is the derivative of (x-1/(2xsqrt(x))) ?
The derivative of (x-1/(2xsqrt(x))) is (-x+3)/(4x^{5/2)}
What is the first derivative of (x-1/(2xsqrt(x))) ?
The first derivative of (x-1/(2xsqrt(x))) is (-x+3)/(4x^{5/2)}